Problem: The lifespans of lions in a particular zoo are normally distributed. The average lion lives $14.4$ years; the standard deviation is $2.5$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a lion living longer than $16.9$ years.
Answer: $14.4$ $11.9$ $16.9$ $9.4$ $19.4$ $6.9$ $21.9$ $68\%$ $16\%$ $16\%$ We know the lifespans are normally distributed with an average lifespan of $14.4$ years. We know the standard deviation is $2.5$ years, so one standard deviation below the mean is $11.9$ years and one standard deviation above the mean is $16.9$ years. Two standard deviations below the mean is $9.4$ years and two standard deviations above the mean is $19.4$ years. Three standard deviations below the mean is $6.9$ years and three standard deviations above the mean is $21.9$ years. We are interested in the probability of a lion living longer than $16.9$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $68\%$ of the lions will have lifespans within 1 standard deviation of the average lifespan. The remaining $32\%$ of the lions will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({16\%})$ will live less than $11.9$ years and the other half $({16\%})$ will live longer than $16.9$ years. The probability of a particular lion living longer than $16.9$ years is ${16\%}$.